Core 3 (C3)

    You can find the formula booklet here and past papers from 2005 to present here. To convert your scores to UMS marks, my online converter tool and Android app are linked from the homepage of
    As you can see in the formula booklet, nearly every formula and result on this page needs to be memorised.

    You do not need to know the numerical methods (decimal search and fixed-point methods) from the coursework for this exam. See the NM module for notes on these methods.


    p1, p2 Types of proof:

    • Proof by direct argument uses a known fact - e.g. that n2n^2 is even if nn is even
    • With proof by exhaustion, you test every possibility. For example, you could prove that there is only one prime number which is 1 less than a square number under 100 by subtracting 1 from all 9 squares under 100 and showing that only one of them is prime
    • Proof by contradiction: for example, if you are trying to prove that 2\sqrt{2} is irrational, you would show that it is false that 2\sqrt{2} is rational. Therefore, if it's not rational, we know that it's irrational
    • With disproof by counterexample, you can show that something is false by giving an example where the conjecture (thing you're trying to prove) fails. For example, you could disprove that all odd numbers under 10 are prime by showing that 9<109 < 10 and 9 is divisible by 3

    Exponentials and natural logarithms

      a1, a2 exe^x:

      • ee is equal to 2.718 to 4 significant figures (it's irrational)
      • On a graph of y=exy = e^x, the gradient is equal to 1 at (0, 1)
      • exe^x is its own derivative; ddx(ex)=ex\frac{\d}{\dx}(e^x) = e^x


      • ln(x)\ln(x) is shorthand for loge(x)\log_e(x)
      • The logarithm laws still apply for ln\ln, e.g. ln(x)+ln(y)=ln(xy)\ln(x) + \ln(y) = \ln(xy) etc

      a3 Graphs:

      • The graphs of exe^x and ln(x)\ln(x) are shown below. The blue is y=exy = e^x and red is y=ln(x)y = \ln(x)
      • The two graphs are inverse functions of each other - they are reflections of each other in y=xy = x
        - This means that ln(ex)=x\ln(e^x) = x, and similarly, eln(x)=xe^{\ln(x)} = x

      a4 Exponential growth and decay:

      • Usually the form y=a×bxy = a \times b^x:
        - If b>1b > 1, it's exponential growth
        - If b<1b < 1, it's exponential decay


        f1 Mapping:

        • Functions map one or more inputs to outputs
        • Functions can either be one-to-one (one input, which outputs one value), or many-to-one. This is because a function cannot produce two values - its purpose is to process a number into another. Therefore, a circle is not a function as any xx coordinate will map to two yy coordinates
        • There are one-to-many and many-to-many mappings, but these are not functions
        • The input is also called the object
        • The output is also called the image

        Domain and range:

        • The set of inputs is the domain (e.g. the xx-coordinates)
        • The set of outputs for the specific inputs used is the range (e.g. the yy-coordinates)
        • To state the type of numbers possible in a domain or range, use the \in symbol followed by the symbol for the type. These are:
          - R\Bbb R: real numbers (any number that isn't the square root of a negative)
          - Z\Bbb Z: integers
          - Q\Bbb Q: rational numbers
          - N\Bbb N: positive integers (excluding 0)
        • For example, the domain of a function could be written as xZ,x3x \in \Bbb Z, x \geq 3, which tells us that xx must be an integer greater than 3
        • Make sure to include this when defining a function - it is not called a function if part of the domain doesn't map to anything

        The codomain:

        • Imagine the function f(x)=x2f(x) = x^2 where xZ,x0x \in \Bbb Z, x \neq 0
        • Because the domain only contains integers (Z\Bbb Z), it will therefore only map to positive square numbers
        • We would write the range as {1,4,9,16,25,...}\{1, 4, 9, 16, 25, ...\}
        • The codomain is the set of values that the range belongs to. Because the range is only positive integers, we would write the codomain as yNy \in \Bbb N

        f1, f8 Odd, even and periodic functions:

        • A function is even if it is symmetrical about the yy-axis; f(x)=f(x)f(-x) = f(x) if f(x)f(x) is even
        • A function is odd if it looks the same when rotated 180° about (0,0)(0, 0); f(x)=f(x)-f(x) = f(-x) if f(x)f(x) is odd
        • Periodic functions repeat periodically (e.g. y=sin(x)y = \sin(x) repeats every 360°)
        • Functions can be more than one of the above, and some are none at all

        f2, f3, f4 Transformations:

        • See C2 for how to apply transformations to f(x)f(x)
        • If you need to do multiple transformations (e.g. 4f(x2)4f(x - 2)), apply one at a time - with this example you could translate it to the right 2 places, and then stretch it ×4 in the yy-axis

        f5 Composite functions:

        • A composite function is a combination of more than one function applied in series, e.g. fg(x)fg(x) is shorthand for the composite function ff[g(x)g(x)]
        • Example question: f(x)=x+1f(x) = x + 1. g(x)=x2g(x) = x^2. Find fg(x)fg(x):
          1) Substitute g(x)g(x) into ff: fg(x)=f(x2)fg(x) = f(x^2)
          2) Now do the same for ff: fg(x)=x2+1fg(x) = x^2 + 1

        f6 Inverse functions:

        • The inverse of f(x)f(x) takes an output of f(x)f(x) and finds the value of xx for it
        • To find an inverse:
          - Write in yy = _ _ _ form
          - Swap xx and yy
          - Solve for yy
          - Replace yy with f1(x)f^{-1}(x)
        • A function f(x)f(x) and its inverse f1(x)f^{-1}(x) are reflections in the line y=xy = x
        • An inverse function must only map to one output, as with a standard function

        f7 arcsin, arccos and arctan:

        • Arcsin is often shortened to sin-1, and the same applies to arccos and arctan
        • The graphs of these functions is the graph of sin, cos or tan reflected in the line y=xy = x (they're inverse functions)
        • For this, the domain must be restricted to create a one-to-one mapping:
          - y=sinxy = \sin x: domain = -90°xx ≤ 90°
          - y=cosxy = \cos x: domain = 0°xx ≤ 180°
          - y=tanxy = \tan x: domain = -90° < xx < 90°

        f9 The modulus function:

        • Modulus (|n||n|) makes all values inside it positive; i.e. |2|=2|2| = 2, |2|=2|-2| = 2
        • Therefore a graph of y=|x|y = |x| never goes below the x-axis, instead reflecting in the line x=0x = 0

        f10 Solving equations with the modulus function:

        • Change the modulus sign to brackets
        • Now solve for xx to find the first solution
        • To find the second solution, add a - sign in front of the brackets and solve again
        • It often helps to sketch a graph
        • The Abs button on most calculators will input the modulus sign, this is useful for checking answers


          c1 The product rule (for u × v):

          • When y=u×vy = u \times v, dydx=u\frac{\dy}{\dx} = u dvdx+vdudx\frac{\d v}{\dx} + v \frac{\d u}{\dx}
          • Example: differentiate (x3+7x1)(5x+2)(x^3+7x-1)(5x+2)
            1) Define uu and vv: u=x3+7x1u = x^3 + 7x - 1, v=5x+2v = 5x+2
            2) u×dvdx=5(x3+7x1)u \times \frac{\d v}{\dx} = 5(x^3+7x-1)
            3) v×dudx=(5x+2)(3x2+7)v \times \frac{\d u}{\dx} = (5x+2)(3x^2+7)
            4) Multiply the two and simplify to 20x3+6x2+70x+920x^3 + 6x^2 + 70x + 9
          • dvdx\frac{\d v}{\d x} is often shortened to vv', and dudx\frac{\d u}{\d x} to uu'

          c2 The quotient rule (for u/v):

          • For differentiating a fraction uv\frac{u}{v}:
            dydx=vdudxudvdxv2\frac{\dy}{\dx} = \frac{v\frac{\d u}{\dx} - u\frac{\d v}{\dx}}{v^2}
          • Order is important; the numerator of the derivative must be vuuvvu'-uv' and not uvvuuv'-vu'

          c3 The chain rule (for composite functions):

          • dydx=dydu×dudx\frac{\dy}{\dx} = \frac{\dy}{\d u} \times \frac{\d u}{\dx}
          • Example: differentiate y=(3x+1)2y = (3x + 1)^2
            1) Differentiate the 2, leaving the inside of the fraction for now:
                dydu=2(3x+1)21=2(3x+1)\frac{\dy}{\d u} = 2(3x + 1)^{2-1} = 2(3x+1)
            2) Differentiate inside the brackets: dudx=3\frac{\d u}{\dx} = 3
            3) Multiply the two: dydx=3×2(3x+1)\frac{\dy}{\dx} = 3 \times 2(3x + 1)
            4) Simplify: dydx=6(3x+1)\frac{\dy}{\dx} = 6(3x + 1)
          • After some practise this can be done by inspection - multiply the power by the inside of the bracket's derivative, then reduce the bracket's power by 1

          c4 Rate of change using the chain rule:

          • dadb\frac{\d a}{\d b} is the rate of change of a with respect to b, how much a changes if b changes by one unit
          • dadb\frac{\d a}{\d b} can always be expressed as a product of two fractions; for example dadc\frac{\d a}{\d c} × dcdb\frac{\d c}{\d b}. Therefore, the chain rule can be used

          An example problem:

          • If a balloon is inflated at a rate of 10 cm310\ \mathrm{cm}^{3} per second, how quickly will the radius increase? (assume that the balloon is spherical)
            1) Write what we know: dVdt=10\frac{\d V}{\d t} = 10 where VV is the volume in cm3\mathrm{cm}^{3} and tt is the time in seconds
            2) Find an equation connecting two variables in the question: V=43πr3V = \frac{4}{3} \pi r^3 (volume of a sphere)
            3) Differentiate this: dVdr=4πr2\frac{\d V}{\d r} = 4\pi r^2
            4) Write what we need to find out: drdt\frac{\d r}{\d t} (how quickly the radius increases with time)
            5) We now have dVdt\frac{\d V}{\d t} and dVdr\frac{\d V}{\d r}. Find an equation: dVdt=dVdr×drdt\frac{\d V}{\d t} = \frac{\d V}{\d r} \times \frac{\d r}{\d t}
            6) Substitute in what we know: 10=4πr2×drdt10 = 4 \pi r^2 \times \frac{\d r}{\d t}
            7) Now rearrange: drdt=104πr2=52πr2\frac{\d r}{\d t} = \frac{10}{4} \pi r^2 = \frac{5}{2} \pi r^2 cm s1\mathrm{cm}\ \mathrm{s}^{-1}

          c5 Differentiating an inverse function:

          • If you know the derivative of a function f(x)f(x), the derivative of f1(x)f^{-1}(x) is 1f(x)\frac{1}{f'(x)}
          • Similarly, 1 over the derivative of the inverse will give the derivative of the original function

          c6 Implicit differentiation:

          • This allows us to differentiate a function without yy being the subject
          • The method uses the chain rule: df(y)dx=df(y)dy×dydx\frac{\d f(y)}{\dx} = \frac{\d f(y)}{\dy} \times \frac{\dy}{\dx}
          • Therefore, you just need to differentiate each yy part with respect to yy and multiply this by dydx\frac{dy}{dx}


          • Differentiate x2+y2+xy=6x^2 + y^2 + xy = 6 with respect to xx:
            2x+2ydydx+y+xdydx=02x + 2y\frac{\dy}{\dx} + y + x\frac{\dy}{\dx} = 0
            Now solve for dydx\frac{\dy}{\dx}:
            2ydydx+xdydx=2xy2y\frac{\dy}{\dx} + x\frac{\dy}{\dx} = -2x - y
            dydx×(2y+x)=2xy\frac{\dy}{\dx} \times (2y + x) = -2x - y
            dydx=2xy2y+x=2x+yx+2y\frac{\dy}{\dx} = \frac{-2x - y}{2y + x} = -\frac{2x + y}{x + 2y}

          c7 Differentiating natural logs:

          • ddx(ex)=ex\frac{\d}{\dx}(e^x) = e^x
          • ddx(eax)=aeax\frac{\d}{\dx}(e^{ax}) = ae^{ax} (you could easily use chain rule instead)
          • ddx(ln(x))=1x\frac{\d}{\dx}(\ln(x)) = \frac{1}{x}
          • ddx(ln(ax))=1x\frac{\d}{\dx}(\ln(ax)) = \frac{1}{x} (because ln(ax)=ln(a)+ln(x)\ln(ax) = \ln(a) + \ln(x) and ln(a)\ln(a) is a constant term which doesn't affect the gradient)

          c8 Differentiating trig functions:

          • ddx(sinx)=cosx\frac{\d}{\dx}(\sin x) = \cos x
          • ddx(cosx)=sinx\frac{\d}{\dx}(\cos x) = -\sin x
          • ddx(tanx)=1cos2x=sec2x\frac{\d}{\dx}(\tan x) = \frac{1}{\cos^2 x} = \sec^2 x

          c9 Integration by substitution:

          • Example: find e4x dx\int e^{-4x}\ \dx
          • 1) Pick a substitution for uu. This is usually given in the exam. Here we'll use u=4xu = -4x
            2) Now differentiate the expression for uu: dudx=4\frac{\d u}{\dx} = -4
            3) Rearrange this to make dx\dx the subject: dx=du4=14 du\dx = \frac{\d u}{-4} = \frac{-1}{4}\ \d u
            4) Now write the integral again with the substitutions: e4x dx=eu14 du\int e^{-4x}\ \dx = \int e^u \frac{-1}{4}\ \d u
            5) This can now be integrated: e4x dx=14eu+c=14e4x+c\int e^{-4x}\ \dx = -\frac{1}{4}e^u + c = -\frac{1}{4}e^{-4x} + c
          • If there is more than one xx, you need to get them all changed to uu (e.g. by rearranging the equation for uu to make xx the subject and then substituting this in)
          • If it is a definite integral (with limits at the top and bottom of the \int symbol), the limits also need to be substituted into the equation for uu
          • If you have a fraction, it should be easiest if you make uu the denominator. If not, it's usually best as the most complex term
          • Remember that you can take constants outside of integrals. For example, 14x dx=14×1x dx\int \frac{1}{4x}\ \dx = \frac{1}{4} \times \int \frac{1}{x}\ \dx

          c10 Integrating 1x\frac{1}{x}:

          • 1x dx=ln|x|+c\int \frac{1}{x}\ \dx = \ln|x| + c, i.e. the natural log of the modulus of xx plus cc

          c11, c12, c13 Other integration results you need to know:

          • eax dx=1aeax+c\int e^{ax}\ \dx = \frac{1}{a}e^{ax} + c
          • sinax dx=1acosax+c\int \sin ax\ \dx = -\frac{1}{a}\cos ax + c
          • cosax dx=1asinax+c\int \cos ax\ \dx = \frac{1}{a}\sin ax + c
          • f(x)f(x) dx=ln|f(x)|+c\int \frac{f'(x)}{f(x)}\ \dx = \ln|f(x)| + c

          c14, c15 Integration by parts:

          • This method can be used when you are integrating two things multiplied together (e.g. xex dx\int xe^x\ \dx). The method is shown below:
            1) Choose uu and vv'. uu should be the simplest part, or lnx\ln x if it contains a natural log. vv' should be the other
            2) Now find uu' by differentiating uu
            3) Find vv by integrating vv'
            4) Use the integration by parts formula udvdx dx=uvvdudx dx\int u\frac{\d v}{\d x}\ \dx = uv - \int v\frac{\d u}{\dx}\ \dx:
            - The left hand side is the integral you're finding
            - The first term on the right is your values for uu and vv multipled together
            - The last term on the right is your values for vv and uu' multiplied together, and then integrated
            5) Make sure that you remember to integrate the last part, it's easy to forget. Also remember +c+ c if it's an indefinite integral